Finite Impulse Response Filters - Digital Signal Processing in Power System Protection and Control - page 68

Fig. 6.1 First five Walsh
impulse functions

1
)

w ( k

0

0

k

-1

0

N − 1

N 21 −

1

w 1 k )

0 k

-1

0

N 21 N −1

−

1

w ( k )

2

0

k

-1

N 41 − N − 1

0

341 N −

1

w 3 k )

0 k

-1

0

N 41 − N 21 −

341 N −

N − 1

1

w 4 k )

0

k

-1

N 81 − 381 N − 581 N − 781 N −

0

N −1

Taking Z transform of ( 6.10 ) one obtains transfer function of the filter:

W 0 ð z Þ¼ X

N 1

z k ¼ 1 z N
1 z 1

ð 6 : 14 Þ

k ¼ 0

and substituting exp ð jX Þ for z one can get the frequency response of the filter:

sin N 2
sin

W 0 ð jX Þ¼ exp j N 1
2

X

2 :

ð 6 : 15 Þ

X

As expected, the filter has linear phase shift versus frequency. It is also seen that
filter gain is equal to N for angular frequency X equal to zero, while the gain
equals zero for frequencies fulfilling equation NX = 2 ¼ mp, and then